Problem: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $18.5$ years; the standard deviation is $4.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living between $22.7$ and $31.1$ years.
$18.5$ $14.3$ $22.7$ $10.1$ $26.9$ $5.9$ $31.1$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $18.5$ years. We know the standard deviation is $4.2$ years, so one standard deviation below the mean is $14.3$ years and one standard deviation above the mean is $22.7$ years. Two standard deviations below the mean is $10.1$ years and two standard deviations above the mean is $26.9$ years. Three standard deviations below the mean is $5.9$ years and three standard deviations above the mean is $31.1$ years. We are interested in the probability of a tiger living between $22.7$ and $31.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the mean. The probability of a particular tiger living between $22.7$ and $31.1$ years is $\color{orange}{15.85\%}$.